Homework Help: First Law of Thermodynamics question

1. Jul 24, 2005

in10sivkid

a quantity of gas in a cylinder recieves 1600J of heat from a hot plate. At the same time 800J of work are done on the gas by outside forces pressing down on a piston. Calculate the change in thermal energy of the gas

Input heat Q = 1600J
The amount of work done on the gas is, W = 800J

now i'm not exactly sure how I'm supposed to set this up

do I use U = Q - W
or Q = W + U

so i either get 800J or 2400J for the answer for U

i know its a simple question but i think i'm just a bit confused conceptually. any help would be great :)

2. Jul 24, 2005

Nylex

It should be $\Delta U = Q + W$, where Q is the heat input to the system and W is the work done on the system. In $\Delta U = Q - W$, W is the work done by the system.

3. Jul 24, 2005

geosonel

you are correct.
it's the first form:
ΔU = Q - W,
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. of course, when work is done to the system (like your problem), W is negative, so that -W is positive, resulting in an increase in internal energy (like your problem):
ΔU = (1600) - (-800) = 2400 J